The ultimate tower of Hanoi algorithm

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There must be millions of people who have heard of the Tower of Hanoi puzzle and the simple algorithm that generates the simplest solution. But what happens when you are playing the game not with three pegs, as in the original puzzle, but with 4, 5, 6 etc. pegs? Hardly anybody seems to know that there are also really really beautiful solutions which are believed to be optimal but whose optimality has only been proved for four pegs. Even less people know that you can boil down all these optimal solutions into simple no-brainer recipes that allow you to effortless execute these solutions from scratch. Clearly a job for the Mathologer. Get ready to dazzle your computer science friends :)
I also talk about 466/885, the Power of Hanoi constant and a number of other Hanoi facts off the beaten track. And the whole thing has a Dr Who hook which is also very cute.
00:00 Intro
01:58 Chapter 1: The doctor vs. the toymaker
14:27 Chapter 2: Hanoi constant
21:21 Chapter 3: The Reve's puzzle
28:04 A beautiful shortest solution for 10 discs and 4 pegs (discs and super-disks)
30:23 Chapter 4: Unprovable algorithm
35:43 A beautiful shortest solution for 10 discs and 5 pegs (discs, super-discs and super-super-discs)
37:17 Supporters
Here are some references for you to check out:
Andreas M. Hinz et al. - The Tower of Hanoi - Myths and Maths, 2nd edition (2018, Birkhäuser Basel) That's the book I mentioned in the video.
The Dr Who episode (well the part that's not been lost) www.dailymotion.com/video/x11...
The 3Blue1Brown video that I mention • Binary, Hanoi and Sier...
Thierry Bousch, La quatrième tour de Hanoi, tinyurl.com/4p3fudu7
That's the paper that pins down things for four pegs.
Andreas M. Hinz, Dudeney and Frame-Stewart Numbers,
A nice paper explaining the connection between Dudeneys's work and Frame-Stewart. Also worth reading for the historical details. tinyurl.com/t8xb2e5t
A. van de Liefvoort, An Iterative Algorithm for the Reve's Puzzle, tinyurl.com/h5cxfy5u
I found this one useful.
Paul K. Stockmeyer, www.cs.wm.edu/~pkstoc/toh.html A couple of very nice papers including a huge bibliography.
Ben Houston & Hassan Masum, Explorations in 4-peg Tower of Hanoi, tinyurl.com/mw95tnek This paper has some pictures of state graphs for the 4-peg puzzle.
towersofhanoi.info/Animate.aspx Very fancy animation of mulit-peg tower of Hanoi. Sadly, it just comes across as a mess of moves for more than three pegs. Programmers, you really should rise to my challenge to animate the 4-peg algorithm the way I present it in this video.
Here is the link to the wiki page for the Celestial toymaker Dr Who episode en.wikipedia.org/wiki/The_Cel...
Makes very interesting reading. Especially the fact that most of this episode has been lost I find pretty amazing. That's also why I only show a still image from the relevant part of the episode and play some audio snippet.
Music: Fresh Fallen Snow and Morning Mandolin both by Chris Haugen, Mumbai effect. All from the free RU-vid audio library.
Enjoy!
Burkard

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1 июн 2024

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Комментарии : 833

@Nikolas_Davis 3 года назад

"If you're ever captured by an evil toymaker, you'll be ready" And they say math has no real-world applications.

@farrankhawaja9856 3 года назад

Underrated comment :)

@warrenbosch3581 3 года назад

Very cool!

@Shadow-xb2ce 3 года назад

I much prefer an evil toymaker to the "motorcycle mathematicians" my sarcastic (as I was told) Calc II teacher used to love throwing us.

@orstorzsok6708 3 года назад

:D :D :D

@titaniumtomato7247 3 года назад

@@Shadow-xb2ce ?

@outputcoupler7819 3 года назад

That's it, I'm stealing that. I'm no longer a Software Engineer, I'm an Algorithmologist.

@Mathologer 3 года назад

Let's hope this word catches on :)

@citolero 3 года назад

17:40 "This recipe also solves another Hanoish puzzle." I think the correct adjective is "Hannoying".

@Mathologer 3 года назад

@StefanReich 3 года назад

Hanoisy

@nickbradley8354 3 года назад

So Hanoid! Saigonna say that. Halong ago you see this? Gotta be quicker next Thai.

@leif1075 3 года назад

@@Mathologer Did you really learn the towers of Hanoi in primary school?? I hope you can PLEASE respond.

@NStripleseven 3 года назад

Hahaha

@henridelagardere264 3 года назад

40 minutes of Professor Polster + links to 400 hours of reading = 1 helluva education Every single time, thank you for all you're doing, Mathologer!

@Mathologer 3 года назад

Enjoy :)

@andrewmartin3671 3 года назад

Hear!

@SaveSoilSaveSoil 3 года назад

I know right? I LOVE the references in the description of Mathologer videos

@l00d3r 3 года назад

I guess it is a BYOW (bring your own wine) party

@ajhokie130 3 года назад

Don't let other RU-vidrs doing the same topics stop you. Please! I like to see different takes on the same topic from my favorite RU-vidrs.

@Mathologer 3 года назад

@superman00001 3 года назад

Hello Mathologer - there is a nice (possibly related, to an extent) puzzle that asks how many different unique ways a spider can put on his shoes and socks, given that each foot must receive a sock followed by a shoe. Can you solve it? (I did, and it’s a nice solution!).

@tcadityaa 3 года назад

Good evening Mr Bukard, I have something to share with you. Our Biology teacher(yes! Our Biology teacher!) told us about an approximation relating Pythagorean triplets. I ran it on Python and its accuracy(in terms of how good the approximation is, or in other words how consistenly low the error percentage is) is very good, though its accuracy rate decreases as we crank up the size. Should we discuss it further here sir? Or how shall i contact you. Because Sir, I assure you, this is not some attention seeking stunt. I trued so hard to figure it out but the 100% accuracy numbers just pop up randomly...I have no idea....

@ChefSalad 3 года назад

You should get yourself a :CueCat as a technological mascot to go with your website. The best part is that it's cat-shaped. It might even still work, if you can find a computer with a PS/2 keyboard port to use it with.

@johnchessant3012 3 года назад

Is it just me or is it very easy to read this paragraph in Mathologer's voice? :)

@Mathologer 3 года назад

@@superman00001 Ah, that's a nice one. Had not heard of this one before :)

@notahotshot 3 года назад

The best thing about Reve's puzzle; "We're going to have some cheese."

@timbeaton5045 3 года назад

But don't do it with camembert on a hot day. You'll end up with one super puddle of cheese.

@heaslyben 3 года назад

Did you wonder if the words "cheeses", "stools", and "moves" were chosen for (juvenile) humor? Perhaps by asking I've revealed too much about my own sense of humor.

@johnchessant3012 3 года назад

13:27 In 1,023 moves, moving the smallest disc every other move means we're moving it 512 times, which is 2 (mod 3). So we need to move it counter-clockwise (A -> C -> B -> A) to make sure it ends at B. In general, 2^(n-1) is 1 (mod 3) if n is odd and 2 (mod 3) if n is even, so we should move the smallest disc clockwise if n is odd and counter-clockwise if n is even.

@zanti4132 3 года назад

As the idea of clockwise vs counterclockwise isn't clear if the three pegs are in a straight line, I think it's easier to remember (a) if n is odd, first move the small disk to the peg that will hold the ending stack (b) if n is even, first move the small disk to the peg that will not hold the ending stack. Then build the smaller towers by alternating between the two pegs until you can move the largest disk to the peg that will hold the ending stack. This essentially reduces the puzzle to solving the tower with n-1 disks. Keep going like this until you are done.

@johannesh7610 3 года назад

I thought about it with the recursion: upper ten to B requires upper nine to C, which requires upper eight to B, 7 to C, 6 B, 5 C,...

@Paul_Hanson 3 года назад

@@zanti4132 In that case, instead of clockwise vs. counter-clockwise it's left and right. And here we hark back to the old video games where if you moved off the right side of the screen, you reappeared on the left side of the screen.

@DerKiesch 3 года назад

@@zanti4132 I personally would find it easier to remember: 1. Odd and even n need a different starting move to end up in the right spot. 2. You check where you end up for the smallest odd stack (1 --> which is only one move). 3. If you have an odd n, it's the same move as in 2, if even n, it's the "opposite" move to 2. 4. After that you just continue moving the small piece in the same "direction" (clockwise / counterclockwise or left/right). Thats an easy way to convince yourself that you have the right direction

@PC_Simo Год назад

@@DerKiesch I agree about that being the easiest way to remember this 👍🏻.

@RaquelFoster 3 года назад

In 1983 I got a Commodore 64 for my 8th birthday. My dad got some 5-1/4" floppy disks full of public domain BASIC games and one of them was Tower of Hanoi. That was the last time I heard of Tower of Hanoi. 🙂

@Mathologer 3 года назад

Pretty sure that was about the same time that I got my Commodore 64. I still remember animating the basic algorithm on this computer. Ah yes, the good old days :)

@RaquelFoster 3 года назад

@@Mathologer Back when animating anything was impressive! I read the Programmer's Reference Guide and was making sprites with a bunch of POKE statements and moving them around the screen. I thought I was going to get hired by Atari as soon as I got the word out! But anything was more advanced than those public domain games. I think they came out of user groups where people just mailed floppies around. I remember a reactor simulator called Nuke 64 which I wanted to like but it was ridiculously simple. But there was a Monopoly game called MONOPOLE-64 that I played a lot. There was some good-looking stuff that only used text mode graphics.

@HansMaurer. 3 года назад

@@Mathologer Did the same with the 80286 PCs that the school had bought for the first ever Grundkurs Informatik (using the recursive algorithm, learned about the circle patterns only later in university).

@Nikolas_Davis 3 года назад

The Tower of Hanoi was the first non-trivial recursion problem I came in contact with, as an example in an introductory textbook of CS in my freshman year at university. I was instantly hooked! :-) None of that boring factorial stuff; now, this was a *real*, powerful algorithm!

@RaquelFoster 3 года назад

@@Nikolas_Davis That's awesome. I don't know if I ever even saw the word recursion in school. I never had any interesting classes. In 1989 I begged my parents for a Turbo Pascal (With Objects! LOL!) compiler for my 14th birthday because I saw it in a store. Then in 1991 I got into a computer camp that had an application with several questions but only one that involved writing code. All you needed to do was find numbers that were both prime numbers and part of the Fibonacci series. I thought it was so easy I needed to make my answer clever to stand out (only 14 kids got into the camp at Ohio Supercomputer Center) so I made a super-small recursive function (not that I knew it was called recursion) that found Fibonacci numbers up to a limit then checked if they were prime as it fell back out. When I got to the camp I asked how they graded it and they said I was the only kid who even got it right LOL. But I got to play with a Cray Y-MP that summer! I hated school so much I kept dropping out because I'd do all my CS work in the first week of class then I'd just have classes I couldn't pay attention in. But one time I kept going until I got an internship before I dropped out. 🙂

@m4riel 3 года назад

After checking a feel cases, I think that to get the tower from A to B, you start by moving the smallest disc to: • C, if there are 2n discs; • B, if there are 2n+1 discs.

@inyobill 3 года назад

Ooooooo, nice thought, I'll have to verify. Well played mate.

@blackmephistopheles2273 3 года назад

You can see that this has to be the case, due to how you move the first (0+1) disk.

@rmsgrey 3 года назад

If you pay attention, you may notice that odd-numbered discs always move one way, while even numbered discs move the other way. Since moving the entire tower involves moving the largest disc once, you want it to move in the right direction. If it has the same parity as the smallest disc, you move them in the same direction; if it's the other parity, they move in opposite directions.

@EebstertheGreat 3 года назад

@@blackmephistopheles2273 You can also work it out by feel as the OP suggests. There are only two possibilities, and they should depend only on the parity of the number of disks, because they depend only on the parity of some permutation in the computation. It all comes down to one choice, and it's deterministic, so either left first means left at the end or left first means right at the end. So you actually only need to check a single case. When there is n=1 disk, you move it onto the target peg. So if there are n=2k+1 disks, make your first move toward the target peg, and if n=2k, toward the other peg.

@ivanskopin7723 3 года назад

When you don't have the corners but you have the pegs, I would phrase it as "if it's an even number, move the smallest disc to your final destination, if it's an odd number, move it to the peg that you don't want the tower to end up on".

@rayhanmansoor2951 3 года назад

So happy to see a new video. I was really upset when you didn't upload in February.

@Mathologer 3 года назад

Yes, sadly insanely busy at uni at the moment. COVID is still messing things up in a big way for us and so hardly any time for Mathologer :(

@niladriroy7650 3 года назад

@@Mathologer 😢

@nonachyourbusiness1164 3 года назад

@@Mathologer How is it affecting you guys? My uni is completely remote now and emphasis has been put on self-study

@Mathologer 3 года назад

@@nonachyourbusiness1164 Well in Australia we seem to have COVID pretty much under control with no new cases out in the open for a month. This means that life is pretty much back to normal, except it isn't. My uni tries to be as much as possible back to face-to-face. Sounds good but is incredibly messy with the government still requiring social distancing which results in all venues only being at half or less capacity. and also things changing all the time in a seismic way as soon as there is the slightest indication that the virus has managed to escape again from hotel quarantine :(

@jacoblojewski8729 3 года назад

If the tower has an odd number of discs, first move the smallest to the place you want the tower to end up. If even, move to the space you don't want the tower to end. Then just consistently follow the direction you moved the smaller disc to (clockwise/counterclockwise if arranged as a triangle, to the left/right with wraparound if arranged in a line).

@peter_p_r_zhang 3 года назад

The toy maker himself wouldn't expect this video to come out one day during that time

@trueriver1950 3 года назад

I think he was a rogue time lord himself. Not only that but both the doctor and the toy maker had probably time travelled to 2021 and seen the Mathologer video...

@TrimutiusToo 3 года назад

I figured out best solution for tower of hanoi when i was 6... My dad gave to me tower of hanoi puzzle with 8 peaces and so that i don't bother him said that i should find best solution and left me like that, and then was very surprised that 3 hours later i came to him and said "255", he already even forgot that he gave me the puzzle, but back in those days i figured out the recursive solution to tower of hanoi which impressed my father quite a bit... I was always good with recursive algorithms...

@Mathologer 3 года назад

Cool :)

@trueriver1950 3 года назад

The iterative algorithm is harder to spot, and would have impressed him even more (I won't give it here, as several other commented already have)

@deffinatalee7699 3 года назад

23:33 there’s another lovely pattern that’s easy to miss in this table; look at the antidiagonals. See anything familiar?

@UCFc1XDsWoHaZmXom2KVxvuA 3 года назад

Increbile, the amount of astractions one's mind can demarcate around a given subject. Just wow, maths is beautiful

@NatePrawdzik 3 года назад

The Toymaker is Hanoi'ing me again.

@eshh183 3 года назад

I remember watching 3B1B's video on Hanoi and feeling such awe, that a seemingly simple puzzle could weave recursion and binary counting together. I never thought anything could replicate that awe again! But damn you! Thank you soo much for making me experience that awe again, and this time much muchh more! I just can't wait to get on my computer and try to animate all the animations that you showed today (and perhaps even a generalised version for small n using Frame-Stewart algorithm?). But seriously lot's of love and respect man for making me fall in love with Maths over and over again!

@Mathologer 3 года назад

That's great. Cannot wait for the first animations to pop up :)

3 года назад

Well, binary counting is recursive just by itself already.

@hybmnzz2658 3 года назад

Honestly 3B1Bs most underrated video

@AAAAAA-gj2di 2 года назад

Don't cuss on a Great Professor who's probably much older than you

@Lightn0x 3 года назад

I used to solve towers of hanoi on fights because I had no internet. Solving the 10-peg version with 1023 moves must've been one of the proudest moments in my life :D Edit: huh, I didn't know the "smallest disk" trick explained at 12:18. I just tried to follow the recursive algorithm in my head. Way to trivialize my life's achievment, Mathologer! /s

@Mathologer 3 года назад

@AngryArmadillo 3 года назад

These are some of your most visually satisfying animations to date, which is really saying something!

@DoReMeDesign 3 года назад

2nd problem is 3^n because there are n discs, each of which has 3 'choices' of pegs - same as the number of vertices on the sierpinski triangle 3rd problem can be done by comparing coefficients of 12/59 and 18sqrt(17)/1003, upon which irrational parts drop out - not sure I want to do this.. -btw, thanks Mathologer!

@HansMaurer. 3 года назад

3^n-1 actually (for the path length)

@Mathologer 3 года назад

@@HansMaurer. That's it :)

@keshavgoel2310 3 года назад

For the second problem, recursion is T(n) = 3T(n-1) +2 as largest disc moves twice and the n-1 disc pile has to be moved from one extreme to another thrice. Solving the recursion gives the promised beautiful answer - 3^n-1

@taibilimunduan 3 года назад

less one, because the initial position does not count as a move

@pisquaredoversix 3 года назад

I have seen a lot of comments before saying that they needed the video before, but I didn't expect this one to be one of those. This was the topic of my research paper for my gifted class last year, specifically the smallest amount of moves to win a k-poled Hanoi Tower, and I plan to carry on this year. I haven't finished the video, but thank you for this wonderful video anyway. Cheers from Korea! Edit: Never thought about 4-peg Hanoi Towers with the concept of superdisks. Thanks again.

@Mathologer 3 года назад

Well, have fun with your research paper :)

@moonblaze2713 3 года назад

Bioware put this in KotOR, and I had to memorize this thankfully simple recipe because I did multiple playthroughs. Years later, they put it in Mass Effect 1 and I still remembered it. And years later I saw this video and still remembered it. Stuff you do repeatedly as a kid really sticks with you, huh?

@inyobill 3 года назад

31:05 I believe I read, was mentioned in one of my Logic classes, or ???: For every Provable True Theorem, there are an infinite number of True Unprovable theorems. "There are more things in heaven and earth, Horatio, than are dreamt of in your philosophy.".

@wompastompa3692 3 года назад

For n disks, the shorthand trick I remember is that odd heights start by moving the small disk to the destination peg and even starts by moving it to the intermediate peg.

@conovan5081 3 года назад

I checked your channel three times this week, I was expecting this, LET'S GOOOOO

@Mathologer 3 года назад

Yes, sadly insanely busy at work and hardly any time for Mathologer :)

@istariknight1 3 года назад

As others have shared, I also really enjoy Mathologer's take on these great topics even if other math channels I follow like 3B1B cover them! Every minute spent on this production by the team was worth its weight in gold (or cheese disks) I always get this sense like it's Christmas morning when I see a new video from you guys, thanks for pumping up my weekend!

@Mathologer 3 года назад

@danielmassart 3 года назад

for the record, T. Bousch also has a strong contender in the "best title for a math paper, ever" category : Le Poisson n'a pas d'arête (Annales de l'Institut Henri Poincaré, 1999)

@NoNameAtAll2 3 года назад

Translate?

@somebodyhere3160 3 года назад

@@NoNameAtAll2 google translate tells me this:the Fish has no bones

@xevira 3 года назад

13:50 if n is odd: clockwise if n is even: counter-clockwise or another way to phrase it: if n is odd: first move is to the target position if n is even: first move is to the OTHER position

@Person1873 3 года назад

The trick I always used for the tower of Hanoi is based on odd or even quantity of discs. If you have an odd number of discs then the first move will be to the target tower. If you have an even number, then your first move will be to the other empty tower.

@denny141196 3 года назад

Decided to implement in python before watching. An interesting point I came across was that when recursing, the midpoint and destination swap for subtower move (so the biggest disc can move to the destination unimpeded). As the algorithm recurses down to the base case of 1 (not 0), there are n-1 such midpoint/destination swaps for a tower of size n. Hence, for towers with an odd number of discs, there are an even number of mid/dst swaps, which all cancel out -> first move should be the first disc on to the intended overall destination. Vice versa for even number of discs.

@buxeessingh2571 3 года назад

For those who don't know: the reason you only saw a still and heard an audio is that only photos from the TV broadcast and a duplicate audio of the telecast are what BBC have in the archives of the first three episodes. Addendum: there was a proposed sequel to "The Celestial Toymaker" called "Nightmare Fair". Rumour had it that an early draft of the script included the Doctor solving the minimal path problem discussed here for the Trilogic game.

@Mathologer 3 года назад

Had not heard about this before :)

@10gbo_pizza 3 года назад

Sir can I say, despite the time, your videos are ABSOLUTELY AMAZING! KEEP UP THE GOOD WORK! :)

@namantenguriya 3 года назад

•Always loving your videos. 🥰🥰🥰 Love from India 🇮🇳🇮🇳🇮🇳. •By the way, if you plug in 3 for n, you get 《946/243》(=3.893) •Please make videos on ☆☆☆ Unsolved problems in Mathematics ☆☆☆.

@Hahahahaaahaahaa 3 года назад

I think, because of how common the Tower puzzle is, and how little I've thought about it before, this is one of the most beautiful videos you've done. Despite many years of studying more complex topics, making "simple" problems elegant often gives me the greatest appreciation of mathematics. Thank you as always!

@tonyalanmarchant7330 3 года назад

I love the little giggles. Could be a bady in a bond movie

@Mathologer 3 года назад

If they ever offer me a role as a mathematical supervillain I'd definitely go for it :)

@tonyalanmarchant7330 3 года назад

@@Mathologer you most defo should.an if the role comes along? I assume you know all about percentages 😁

@sidimohamedbenelmalih7133 3 года назад

Yes mathologer video when i'm bored is the best of the best

@Gravitraxer_AangCZ 3 года назад

It is indeed!

@blackmephistopheles2273 3 года назад

#1: Nine disks need forty-one (41) moves. My solution is the "minimum of (two times a smaller number of disks plus the minimum of the remaining disk moves) method. #2: An observation: if d (# of disks) < p (# of pegs), then you ALWAYS only need 2*d-1 moves. Basically, having more pegs than disks means you can move every disk to its own peg, including the base disk (to the designated ending peg), then move all of the other disks back on to the next bigger disk, which you should have already moved to that ending peg: (d - 1) + 1 + (d - 1) = 2d - 1. #3: Likewise, if d = p, then m = 2d + 1, since you will have to move whichever disk is on the ending peg (or move another disk, if the ending peg is the only one available to the biggest disk), move the biggie, move the disk you had to move beforehand, and then move all the other disks onto the biggie: (d - 1) + 1(clear off the ending peg) + 1(biggie) + 1(split the double stack up with the starting peg) + (d - 1) = 2d + 1.

@walrusman8691 3 года назад

This kind of single move algorithms often show up in computer science and an often overlooked follow up question is if their were "multithreading" where if two subsequent movements occurred to and from different pegs they could be done simultaneously as 1 move. Obviously this would only apply to larger n but I would be curious if the solutions for that were actually superior given how ordered the solutions seem to be already.

@heaslyben 3 года назад

I love that question. It makes me think of context switching as well. Could any process (aka intelligent but forgetful prisoner) wake up, inspect the current state of the board, and make the next correct move? My instinct is "yes" but I'm not sure.

@aeleron0577 3 года назад

@@heaslyben probably. Of course it depends on the actual problem, but one could use a hamming code approach for that. Aka coding a failsafe

@MatejCrhak 2 года назад

I remember finding formula for the smallest number of moves required for n disks and an algorithm to solve it, when I were 12 years old. Those were great times.

@KillianDefaoite 3 года назад

Me every time a Mathologer video comes out: *visible excitement*

@hamiltonianpathondodecahed5236 3 года назад

"Oh , I know about this puzzle" "how tf I never thought about multiple pegs"

@seanofpeace 3 года назад

While I never faced it, I was considering the programming interview question: "Solve the Tower of Hanoi puzzle using a recursive function." Here's my problem with this interview question: It doesn't actually *need* a recursive function. I'm of the opinion that if you can craft more supportable code, however less "elegant," it is, you will be doing right by your client. Here's how I would structure it (using English rules for pseudocode): Assume N disks (Labeled Disk 1 through Disk N) Assume 3 rods (Labeled 1, 2, and 3) Assume all disks are first placed on Rod 1. All moves follow the rules (The only disk that can ever be moved to either rod is Disk 1. All other exposed disks can only move to the rod not occupied by Disk 1.) Rule 0: (One time move) 　If N is even, move Disk 1 to Rod 2; 　　Else, move Disk 1 to Rod 3. LOOP ((N^2)-2)/2 times: 　Step 1: 　　Move the smallest disk that is not Disk 1. 　Step 2: 　　If Disk 1 is not on Disk 2, move Disk 1 to Disk 2; 　　　Else, 　　　If the two non-1 disks are both even or both odd 　　　　Move Disk 1 to cover the larger disk 　　　　Else, 　　　　Move Disk 1 to cover the even disk That's it. One single one-time move, and a loop that bounces back and forth between a non-decisive move, and a 3-option decision move. I believe this approach would be easier to read and maintain by someone who came along later. "Elegance" is great if it saves time, and is clear and easy to maintain. But it doesn't have client value in and of itself.

@maxsch.6555 3 года назад

Oh, what could be nicer than a weekend with a mathologer video :)

@PapaFlammy69 3 года назад

Let's gooooooooooooooooooooooooo! :D

@Mathologer 3 года назад

Guten Abend Jens :)

@meeeshroom 3 года назад

Papa Flammy ist hier!

@maxwellsequation4887 3 года назад

Yaaaaaaaaaaaayyyyyyyyyy

@adityaruplaha 3 года назад

Hello papa

@PapaFlammy69 3 года назад

@@Mathologer Ich hoffe du hattest ein schönes Wochenende bis dato :3

@PC_Simo Год назад

18:20 The number of moves is just the number of possible states (3^n) minus 1, since the algorithm visits each state exactly once; and since the initial state requires 0 moves to reach from the initial state. The number of pegs is always 3; so, the number of moves, for n discs, is: (3^n)-1. Tada! 😎

@AlexanderQ689 3 года назад

Excellent video, I love the visuals & extending the tower of hanoi puzzle beyond what's typically talked about

@AmitSingh-sf5qp 3 года назад

My interest in maths came back after my exam .

@stromboli183 3 года назад

Easily one of the best channels on all of youtube. Thank you once again for a very interesting, entertaining, learnful, and inspiring video, professor Burkard! 🙌

@abrvalg321 3 года назад

0:56 that puzzle becomes really hard really quick if allow to input not only number disks but sticks as well.

@danpost5651 3 года назад

14:00 For odd number of disks, move clockwise; for even, move counter-clockwise. Easy to see. if one disk (or odd), move directly to ending peg; otherwise, not.

@HAL-oj4jb 3 года назад

I didn't even notice that the movie and tv references were gone, nice to see a revival!

@trueriver1950 3 года назад

Try opening the pod bay doors...

@alvarezjulio3800 3 года назад

Once time again a Wonderful Video from Mathologer !!! Thank you.

@leonidassavalas745 3 года назад

There was no Easter egg at the end, but I always watch the entire (always wonderful) video(s). Thank you, Mathologer!

@Mathologer 3 года назад

That's great :)

@TheJoaovascorodrigue 3 года назад

Your videos are mind-ticklers! Thank you

@davidrosa9670 3 года назад

30:07 Minimum number of moves for 9 disk and 4 pegs (I think): 41 (^.^). You explained how to divide the disks into 3 superdisks for 8 disks but not for 9. So, just for fun, I tried every possible division in a very inefficient c++ code. I found that you can solve optimally in 41 moves if you divide the disks into 3 superdisks of 2, 3 and 4 disks (from top to down); also dividing them into 4 superdisks of 1, 1, 3 and 4; 1, 2, 2 and 4; AND 1, 2, 3 and 3. I didn't check if these divisions provide different solutions (Challenge for anybody?), but my guess is that, at least, the 3 and 4 superdisk divisions are different optimal solutions. Calculation (hopefully) explained: Assume that you got to solve for 8 disks and 4 pegs, and you divide the 8 disks into 3 superdisks of 1, 3 and 4 disks (from top to down, like Mathologer did). Also note that f(n)=2^n-1 is the optimal number of moves to solve n disks and 3 pegs. Then, (hopefully) it's easy to see that the number of moves, following the Mathologer path, is equal to f(1)+f(3)+f(1)+f(4)+f(1)+f(3)+f(1)=4f(1)+2f(3)+f(4)=2^2 f(1)+2^1 f(3)+2^0 f(4). Note that the arguments of the f's add up to the 8 disks and also note the decreasing powers of 2. (hopefully) it's easy to see that, for n disks, 4 pegs and m superdisks of a1, a2, ..., am disks, such that a1+a2+...+am=n, the resulting number of moves is equal to 2^(m-1)f(a1)+2^(m-2)f(a2)+...+2^0 f(am). For the code, I simply apply this formula to every possible division of the 9 disks into superdisks. I cycle through the possible divisions in a very inefficient and lazy way: I go though all possible lists of between 1 and 9 elements with all possible values between 1 and 9, checking if the accumulation of the elements add up to 9. Seeing the formula, it's easy to see that it's not worth it to check superdisk divisions that are not in decreasing amounts of disks, but whatever. Here's the code for anyone interested: #include #include #include //accumulate algorithm using namespace std; int pow(int b,int e){//b to the power of e with e being a non negative integer int ans=1; for(int i=0;i

@cybersecuritytutorials8762 2 года назад

Amazing!!! Enjoy your storytelling style to explain the mathematical concepts and algorithms.

@Jukka70 3 года назад

So happy to see the video for this month. Wish they were more times then once a month, but am sure it takes a long time to make these, and happy that you make them

@Mathologer 3 года назад

Yes, sadly my uni job keeps me insanely busy these days. That and family leaves hardly any time for Mathologer. Anyway, I'll keep trying to put one out every month.... :)

@Veptis 3 года назад

It's often used to teach recursion. And last semester our professor used Fibonacci sequence and towers of Hanoi as two examples to show different programming languages. Including postscript, which was quite fascinating

@cameronlakeview7068 3 года назад

I'm currently finding tower of Hanoi a good muse problem for getting my flat in order! (Long overdue tidying) So extra specially well pleased I watched this recently! (I'm picking things up off one surface, and putting them down on another temporarily.. Tower of Hanoi, I've learned (through this video) teaches that it's much less work with more permanent (rather than temporary - eg a table top or 2, rather than balancing pins on the tops of empty beer bottles while waiting to see if enough needless etc will emerge from the mess quantity to warrant a "sewing stuff box"...

@jorgevegarevolledo 3 года назад

Todos tus videos son espectaculares. Gracias por tomarte el trabajo de hacerlos, son realmente inspiradores.

@_majortom_ 3 года назад

beautifuly done as always. I have enjoyed every part of this video. thanx!

@stulora3172 3 года назад

15:52 I love the Michael Ende quote

@stulora3172 3 года назад

unter uns.... war das Absicht?

@notabotta3901 3 года назад

Couldn’t be happier to see this video and try the challenges!

@melekhine 3 года назад

Every one of your videos is so amazing, Professor Burkard. Best on the internets

@alaksiejstankievicx 3 года назад

I've paused on 5:04, because my answer is YES. I'm definitely sure, that I will do this exactly in 1023 moves. Course I read a lot about Hanoi puzzle, couple of time programmed it and taught other people to program it on the lectures as exercises on recursion, but only last year I bought my own physical version (with 8 discs), what leads me to so brave assertion. Only after couple of times "physical/manual" solution I really grocked key principles needed to solve this in circumstance similar to Dr. Who's. First one is really mentioned recursion but in not very computer form rather intention: you want to move subtower on n-1 discs on auxiliary peg to be able to move the disc just under subtower (n-th) to destination peg. Second and this is very important for us human to not solve recursion every time: if your subtower (including full tower) consists of odd number of discs you very first move should be on destination (for current subtower move) peg, if your subtower consists of even number of discs you very first move should be on auxiliary peg. Having this principles you will be able to move Hanoi tower as a calming exercise/solitaire as I do. Now I will continue to watch your video to find what is Mathlogger solution.

@alaksiejstankievicx 3 года назад

12:36 And it seems (of course the solution is the same) your algorithm especially in human perception differs from my. Interesting.

@manorueda1432 3 года назад

Funny and awesome, as always! I've enjoyed it. I'll think about the puzzle with for and five pegs!

@chiranjivipokharel1152 3 года назад

Always so curious to see vedios , thanks a lot mythologer , love from Nepal

@awesomegamedev 3 года назад

I've been recently working on Tower of Hanoi animations for my channel and in the process discovered some of the stuff mentioned in the video, for example the algorithm to solve Tower of Hanoi without recursion by moving the smallest disk every second turn. I also proved why it works. Answer to challenge questions: Direction to move the smallest disk is: if (n is even) then Right else Left That's because to move (N)-tower in some direction, you first have to move (N-1)-tower in the opposite direction, so the direction to move the smallest disk changes every time and it starts with Left for N=1 disk The number of turns is (as mentioned in the video) equal to the number of reachable states, which is = 3^n, because: there are 'n' disks, each disk can be placed in one of 3 towers, so there are 3^n valid states. Each valid state is reachable with a standard recursive algorithm, so there are 3^n reachable states. Another interesting puzzle for you: given a number of disks (N) and a number of turns (T), how do you quickly find the state of all disks in the optimal solution of N disks after making T turns? I had to solve this puzzle while making my Tower of Hanoi animations, so I can instantly preview my animation at any moment in time (instead of re-playing it from the start), which becomes increasingly important with several-hour animations. Check out my channel for these animations: I've already published the shorter ones and longer once, including 10-hour 16-disk solution animation, coming soon:)

@josephcalzaretta8516 3 года назад

This was so cool that I couldn't stop thinking about it. 30:14 "For my programmer friends..." I wrote a demo you can see at jcalz.github.io/hanoi-demo/ that (I think) implements and animates the Frame-Stewart algorithm. Last time I posted this I think it got sent to the spam graveyard, so hopefully his one will stick.

@Muhahahahaz 6 месяцев назад

Awesome work! I immediately put in 5 pegs with 100 disks, and surprisingly it didn’t break lol (I set it to a speed of 0.01s so that I could see the whole thing in under a minute)

@Muhahahahaz 6 месяцев назад

I also just tried 100 disks with 101 pegs (which is the minimal number of pegs required to minimize disk movement - it only requires 199 moves, which is 2 moves for every disk except the largest one) Much faster than 5 pegs! But obviously not very interesting lol

@troemax 3 года назад

Wonderful video. Thank you for making these!

@laierr 3 года назад

I'm so glad to see (literally) that your color scheme became more colorblind-friendly!

@yonatan1myers 3 года назад

for the Q in 14:00: if mod(n,2) == 0 start by moving the smallest disk to the peg that is not your target, else start by moving it to the target

@sidazhou 3 года назад

Already smashed the like button multiple times before Intro was finished. Such was the enthusiasm to learn.

@Mathologer 3 года назад

That's great :)

@AlessandroDruetto 6 месяцев назад

18:16 (shortest solution with extra adjacent-peg rule): it is obvious that we have to visit all the nodes in the graph, and a path traversing x nodes exactly once requires x-1 arcs. So, the question boils down to evaluate the number of nodes in such a graph, that is essentially composed by all the possible valid configurations of disks on pegs. With 1 disk we have 3 valid configurations (the smallest triangle), with 2 disks we have 9 valid configurations (the 2nd-smallest triangle), with 3 disks we have 27 valid configurations (...), etc. Note that when we assign a "set" of disks to a peg, their placement is uniquely determined (since it must comply the no-big-over-small rule) by the total ordering of disk sizes. Then, the total number of configurations is the number of ways to assign n disks to 3 pegs; that is the question "with n available digits, how many numbers can I build that are 3-digits long" and the answer is 3^n. Having 2^n nodes in the graph, having to traverse all of them exactly once, the shortest "adjacent-peg" solution is of length 3^n-1 with 3 pegs; more generally with p pegs the shortest "adjacent-peg" solution is of length p^n-1 by the same reasoning.

@xCorvus7x 3 года назад

13:27 The doctor has to move the tip counter-clockwise because B in that picture corresponds to the lower right corner of the triangle we had looked at before. In general, if you have a tower with n disks, you have to move the tip clockwise for odd n and counter-clockwise for even n for the tower to end up at B (according to this picture). Moving a tower with an even number of disks to B means that you first move a tower with one less disk and thus an odd number of disks to C and since a single-disk tower can be moved with one move only, the tip of a tower with an odd number of disks moves initially to the place where the whole tower will eventually end up. Therefore, if the tower has an even number of disks, the tip has to move to a different place than the final location.

@vivekpal1004 3 года назад

Such a great explanation! Excellent!!!

@maxreenoch1661 3 года назад

Mathologer: *Explains mathematical induction in under a minute* Teachers: *take notes furiously*

@trenthawkins Год назад

I've played enough (3 peg) Hanoi puzzles in various video games that moving the tower in the minimum moves is easy... I sometimes end up second guessing myself on the 'direction' the tip of the tower has to go... Although seeing the pattern play out, I now get that you only have to determine which direction the base has to move, and then the direction the top piece moves is based on if there are an odd number of pieces (same direction as the bottom), or even (opposite direction of the bottom).

@ripe_aces 3 года назад

14:43, this wicked sense of humor had me in stitches.

@StankyPickle1 3 года назад

Beautiful visualizations!

@asadshahpasha 3 года назад

Video uploaded 1 minute ago and already had 5 likes. I am really impressed by the amazing mind power of mathologer's viewers where they fast forwarded a 40 minute video to watch in 1 minute and still understood it well to like the video at the end. Or may be they used some kind of algorithm i am mot sure. Great job guys!

@Mathologer 3 года назад

Yes, amazing isn't it :)

@trueriver1950 3 года назад

Dr Who and the Toy Maker both had time machines. I bet they were the first two likes Or on reflection perhaps not the Doctor as his TARDIS was notoriously erratic at teaching the intended destination

@woodpecker8546 3 года назад

What a fantastic video again! Thank you very much!

@stingrayx2182 3 года назад

One of your best videos, thanks!

@trtlphnx 3 года назад

Love Your Mind, My Friend,,, Keep up the Incredible Work; You have a #1 Fan Here!!!

@paulklee5790 3 года назад

I distinctly remember seeing this on our steam driven television the first time round... by then I’d graduated from watching from behind the sofa (for safety reasons...), it had a big impact on me ... I made a version of the game at school and started a craze for playing it that lasted two whole weeks! Nerd heaven.....

@matthewdodd1262 3 года назад

13:55 Answer is pretty simple. If N is odd, you move it clockwise to end on B, and counterclockwise to end on C If N is even, you move it clockwise to end on C, qnd counterclockwise to end on B. It has to do with where disc 2 and disc 3 land, and proving that its to do with odd-even parity, and that it holds for all N

@johnchessant3012 3 года назад

Brilliant and very satisfying video!

@samuelwaller4924 3 года назад

I just recently found a code golf where the challenge was to quine that output a unique program every iteration, without repeat, for however many iterations. I immediately thought of Gray code, and have been absorbed. This is great!

@ryanbell3704 3 года назад

18:10 I believe the answer is just 3^n-1, since that’s how many possibly positions there are (3 choices for each disc, so 3^n possibly positions, so 3^n-1 moves to get through all of them) really neat how that only-adjacent rule gets us from 2^n-1 to 3^n-1

@surfing_youtube 2 года назад

Thank you for this excellent video. It was worth watching.

@keshavgoel2310 3 года назад

@benYaakov 3 года назад

Very nice , keep sharing such knowledge at RU-vid platform :)

@Tiessie 3 года назад

Your videos are so interesting and inspiring

@prashantshukla6018 3 года назад

Oh gosh I was craving such stuffs 😂. Love ur creations sir huge love and respect from India . Thank u sir!

@SaveSoilSaveSoil 2 года назад

I find myself revisiting your hugely helpful references again. Kudos to Team Mathologer!

@Mathologer 2 года назад

Glad to hear it!

@Joker22593 3 года назад

Ah, Dynamic Programming! Excellent.

@d4django 3 года назад

I am ready for the tomaker now! Thankyou Mr Mathologer

@dcterr1 3 года назад

By the way, Dr. Who was a great show! Great choice for introducing the Tower of Hanoi puzzle!

@fnln-namaemyouji 3 года назад

I had the Tower of Hanoi on my graphing calculator way back in middle school, and used to 'speedrun' 10 disc solutions when I was bored in math class, so I can actually say with relatively high confidence I could still pull it off without a mistake. There's a nice rhythm to it once you get the hang of it, almost meditative. Interestingly, because it was presented as a straight line of pegs instead of a triangle of them, I never stumbled on the clockwise/counterclockwise solution to figuring out where the top disc goes, another example of how how you frame a problem affects how you solve it. Instead, my method was to think in terms of a start peg, a storage peg, a goal peg, and partial pyramids. Move the top part of the pyramid to one peg (the target peg if the pyramid has odd number of discs, the storage peg if it has an even number), move the top disc of the bottom part of the pyramid to the other peg, move the top part of the pyramid on top of it, and now you have a bigger top part and a smaller top part. Repeat until there's just one disc on the start peg, move it to the other peg, and then you have a new whole pyramid, and you recurse back to the start, just with the start and the storage peg switched places, and a pyramid one disc shorter.

@Mathologer 3 года назад

The tower of Hanoi was the first animation I did on a computer (way back in the stone age :)